Which relation on A={1,2,3} is transitive but not symmetric?

Correct answer: R={(1,1),(2,2),(1,2)}

Correct answer: A. A relation is transitive when, whenever (x,y) and (y,z) belong to it, (x,z) must also belong to it. It is symmetric when (x,y) implies (y,x). In option A, (1,2) is present but (2,1) is absent, so the relation is not symmetric. It is transitive because the only possible chains either involve a diagonal pair, which leaves the other pair unchanged, or use (1,2) followed by (2,2), which again gives (1,2), already present. Thus A satisfies both required conditions. Option B is symmetric because it contains both (1,2) and (2,1), but it is not transitive: these two pairs would require (1,1), which is absent. Option C is not transitive because (1,2) and (2,3) are present but (1,3) is missing; it is also not symmetric. Option D is not transitive: (2,1) and (1,2) would require (2,2), which is absent. The key test is to search for a two-step chain and then check the required shortcut pair.